Exercise 4 Clustering and classification.
#Access the MASS package.
library(MASS)#Load the data.
data("Boston")#Explore the dataset.
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00#Plot matrix of the variables.
pairs(Boston)
There are 506 observations (rows) and 14 variables (columns) in the dataset. The variables are for example crim (per capita crime rate by town), zn (proportion of residential land zoned for lots over 25,000 sq.ft), indus (proportion of non-retail business acres per town), chas (Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)), nox (nitrogen oxides concentration (parts per 10 million)), lstat (lower status of the population (percent)) and medv (median value of owner-occupied homes in $1000s).
For more details of the dataset, please visit https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:MASS':
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## select## The following objects are masked from 'package:stats':
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## filter, lag## The following objects are masked from 'package:base':
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## intersect, setdiff, setequal, unionlibrary(corrplot)## corrplot 0.84 loaded#Calculate the correlation matrix and round it.
cor_matrix<-cor(Boston) %>% round(digits = 2)#Print the correlation matrix.
cor_matrix## crim zn indus chas nox rm age dis rad tax ptratio
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58 0.29
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31 -0.39
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72 0.38
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04 -0.12
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67 0.19
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29 -0.36
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51 0.26
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53 -0.23
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91 0.46
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00 0.46
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46 1.00
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44 -0.18
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54 0.37
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47 -0.51
## black lstat medv
## crim -0.39 0.46 -0.39
## zn 0.18 -0.41 0.36
## indus -0.36 0.60 -0.48
## chas 0.05 -0.05 0.18
## nox -0.38 0.59 -0.43
## rm 0.13 -0.61 0.70
## age -0.27 0.60 -0.38
## dis 0.29 -0.50 0.25
## rad -0.44 0.49 -0.38
## tax -0.44 0.54 -0.47
## ptratio -0.18 0.37 -0.51
## black 1.00 -0.37 0.33
## lstat -0.37 1.00 -0.74
## medv 0.33 -0.74 1.00#Visualize the correlation matrix.
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)
Positive correlations are displayed in blue and negative correlations in red color. Color intensity and the size of the circle are proportional to the correlation coefficients. There is a high negative correlation between indus (proportion of non-retail business acres per town) and dis (weighted mean of distances to five Boston employment centres), nox (nitrogen oxides concentration (parts per 10 million)) and dis (weighted mean of distances to five Boston employment centres), age (proportion of owner-occupied units built prior to 1940) and dis (weighted mean of distances to five Boston employment centres) and istat (lower status of the population (percent)) and medv (median value of owner-occupied homes in $1000s). There is a high positive correlation between rad (index of accessibility to radial highways) and tax (full-value property-tax rate per $10,000).
Standardize the dataset, create a categorical variable of the crime rate, divide the dataset to train and test sets.
#Center and standardize variables.
boston_scaled <- scale(Boston)#Summaries of the scaled variables.
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865#Class of the boston_scaled object.
class(boston_scaled)## [1] "matrix" "array"#Change the object to data frame.
boston_scaled <- as.data.frame(boston_scaled)After scaling the mean is 0 for all the variables which means that all variables are normally distributed.
#Summary of the scaled crime rate.
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110#Create a quantile vector of crim and print it.
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610#Create a categorical variable 'crime'.
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))#Look at the table of the new factor crime.
table(crime)## crime
## low med_low med_high high
## 127 126 126 127#Remove original crim from the dataset.
boston_scaled <- dplyr::select(boston_scaled, -crim)#Add the new categorical value to scaled data.
boston_scaled <- data.frame(boston_scaled, crime)Divide the dataset to train and test sets, so that 80% of the data belongs to the train set.
#Number of rows in the Boston dataset.
n <- nrow(boston_scaled)#Choose randomly 80% of the rows.
ind <- sample(n, size = n * 0.8)#Create train set.
train <- boston_scaled[ind,]#Create test set.
test <- boston_scaled[-ind,]#Save the correct classes from test data.
correct_classes <- test$crime#Remove the crime variable from test data.
test <- dplyr::select(test, -crime)Fit the linear discriminant analysis on the train set.
#Linear discriminant analysis.
lda.fit <- lda(crime ~ ., data = train)#Print the lda.fit object.
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2772277 0.2450495 0.2351485 0.2425743
##
## Group means:
## zn indus chas nox rm age
## low 0.9945094 -0.9201932 -0.09656566 -0.8659183 0.47912637 -0.8863873
## med_low -0.1204025 -0.3368662 -0.03371693 -0.5733067 -0.07193602 -0.3175811
## med_high -0.3771136 0.2119824 0.14210254 0.4960739 0.07511365 0.4189697
## high -0.4872402 1.0149946 -0.03128211 1.0434019 -0.39767020 0.8171062
## dis rad tax ptratio black lstat
## low 0.8784783 -0.6927036 -0.7565926 -0.46652346 0.37590407 -0.77863709
## med_low 0.3132539 -0.5375653 -0.5001733 -0.06857581 0.32516802 -0.19739767
## med_high -0.4063095 -0.4499496 -0.3057599 -0.33828828 0.06103381 0.01083282
## high -0.8559398 1.6596029 1.5294129 0.80577843 -0.87606523 0.90851382
## medv
## low 0.55969257
## med_low 0.03992492
## med_high 0.14888371
## high -0.66360176
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.07243097 0.60309925 -0.98002993
## indus 0.16147115 -0.21907704 0.23410171
## chas -0.10269437 0.01482725 0.13361394
## nox 0.10609962 -0.87903847 -1.50524971
## rm -0.09856199 -0.08908100 -0.13838146
## age 0.24522013 -0.28643227 -0.01873592
## dis -0.06009328 -0.25424831 -0.01953038
## rad 3.89268768 1.24980925 -0.09315176
## tax 0.24084338 -0.28421884 0.55726625
## ptratio 0.10995485 0.01518102 -0.29336261
## black -0.13374676 0.03197006 0.14973297
## lstat 0.22107063 -0.28124672 0.24837648
## medv 0.23417664 -0.39539808 -0.24266221
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9598 0.0302 0.0100#The function for lda biplot arrows.
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}#Target classes as numeric.
classes <- as.numeric(train$crime)#Plot the lda results.
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
The best linear separator is variable index of accessibility to radial highways (rad), which has the longest arrow in the picture. The second best separator is difficult to make out, but it looks like it would be nitrogen oxides concentration (parts per 10 million) (nox).
Predicting with the model.
#Predict classes with test data.
lda.pred <- predict(lda.fit, newdata = test)
#Cross tabulate the results.
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 9 6 0 0
## med_low 6 15 6 0
## med_high 1 13 14 3
## high 0 0 1 28Note that the numbers change every time you run the model. The model predicted the high crime rates well (32/32 were classified correctly). Other categories the model did not predict as well: For medium high crime rates, 17/29 were classified correctly, for medium low crime rates 17/25 were classified correctly, and for low crime rates 9/16 were classified correctly.
total <- c(9+5+2+3+17+5+10+17+2+32)
total## [1] 102correct <- c(9+17+17+32)
correct## [1] 75Out of a total of 102 observations, 75 observations were classified correctly.
ratio <- c(correct/total)
ratio## [1] 0.7352941Accuracy of the model was 74%, which is not that bad but could be better.
Reload Boston dataset.
library(MASS)
data("Boston")#Center and standardize variables.
boston_scaled <- scale(Boston)#Change the object to data frame from matrix type.
boston_scaled <- as.data.frame(boston_scaled)#Calculate the Euclidean distances between observations.
dist_eu <- dist(boston_scaled)#Look at the summary of the distances.
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970Run k-means algorithm on the dataset.
#K-means clustering.
km <-kmeans(boston_scaled, centers = 3)#Plot the Boston dataset with clusters.
pairs(boston_scaled, col = km$cluster)
#Investigate the optimal number of clusters and run the algorithm again.
set.seed(123)#Determine the number of clusters.
k_max <- 10#Calculate the total within sum of squares.
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})#Visualize the results with qplot. Visualize (with qplot) the total WCSS when the number of cluster goes from 1 to 10.
library(ggplot2)
qplot(x = 1:k_max, y = twcss, geom = 'line')
2 clusters seems optimal as the bend (knee) is at 2.
#Run kmeans() again with two clusters.
km <-kmeans(boston_scaled, centers = 2)#Plot the Boston dataset with clusters.
pairs(boston_scaled, col = km$cluster)
Like observed before, the optimal number of clusters seems to be two.
Super bonus.
model_predictors <- dplyr::select(train, -crime)#Check the dimensions.
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3#Matrix multiplication.
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)#Matrix multiplication.
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:ggplot2':
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## last_plot## The following object is masked from 'package:MASS':
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## select## The following object is masked from 'package:stats':
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## filter## The following object is masked from 'package:graphics':
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## layout#Create a 3D plot of the columns of the matrix product by typing the code below.
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.#Add argument color as an argument in the plot_ly() function.
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)Draw another 3D plot where the color is defined by the clusters of the k-means.
#Make a k-means with 4 clusters to compare the methods.
km3D <-kmeans(boston_scaled, centers = 4)plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = km3D$cluster[ind])Medium high crime rates seems to bee better defined than cluster 1. Cluster 2 is a bit better defined (not so much intermingling) than low crime rate in the first picture. Cluster 3 is quite similar with high crime rates, even though cluster 3 is a bit more better defined. Cluster 4 is more defined than medium low crime rates in the first picture.